Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Bipolar Junction Transistors
• Physical Structure & Symbols
• NPN
Emitter
(E)
n-type
Emitter
region
p-type
Base
region
Base
(B)
Emitter-base
junction (EBJ)
C
n-type
Collector
region
Collector
(C)
B
Collector-base
junction
(CBJ)
E
(a)
(b)
• PNP - similar, but:
• N- and P-type regions interchanged
• Arrow on symbol reversed
• Operating Modes
Operating mode
EBJ
CBJ
Cut-off
Reverse
Reverse
Active
Forward
Reverse
Saturation
Forward
Forward
Reverse-active
Reverse
Forward
• Active Mode - voltage polarities for NPN
IC
VCB > 0
C
B
IB
VBE > 0
E
IE
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
BJT - Operation in Active Mode
p
n
{
E
IE
IEn
electrons
IEp
holes
n
C
IC
recombination
IB
B
• IEn , IEp both proportional to exp(VBE/VT)
• IC ≈ IEn
⇒ IC ≈ IS exp(VBE/VT)
(1.1)
• IB ≈ IEp << IEn
IC = β IB where β large
⇒ can write
• IS = SATURATION CURRENT (typ 10
-15
to 10
(1.2)
-12
A)
• VT = THERMAL VOLTAGE = kT/e ≈ 25 mV at 25 °C
• β = COMMON-EMITTER CURRENT GAIN (typ 50 to 250)
• Active Mode Circuit Model
IB
IC
B
C
β IB
IE = IB + I C
E
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
BJT Operating Curves - 1
IC vs VBE
• INPUT-OUTPUT
(for IS = 10
-13
A)
IC (mA)
100
IC
80
V CB > 0
ACTIVE
CUT-OFF
60
C
B
40
VBE
E
20
VBE (V)
0.0
0.2
0.4
0.6
0.8
• ACTIVE REGION:
• IC ≈ 0 for VBE < ≈ 0.5 V
• IC rises very steeply for VBE > ≈ 0.5 V
• VBE ≈ 0.7 V over most of useful IC range
• IB vs VBE similar, but current reduced by factor β
• CUT-OFF REGION:
• IC ≈ 0
• Also IB , IE ≈ 0
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
BJT Operating Curves - 2
(for β = 50)
IC vs VCE
• OUTPUT
IC (mA)
12 SAT
ACTIVE
10
IB = 200 µA
IC
C
IB = 160 µA
8
IB = 120 µA
6
IB = 80 µA
4
B
IB
VCE
E
IB = 40 µA
2
VCE (V)
0
0
1
2
• ACTIVE REGION (VCE > VBE):
• IC = β IB , regardless of VCE
i.e.
CONTROLLED CURRENT SOURCE
• SATURATION REGION (VCE < VBE):
• IC falls off as VCE → 0
• VCEsat ≈ 0.2 V on steep part of each curve
• In both cases:
• VBE ≈ 0.7 V if IB non-negligible
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Summary of BJT Characteristics
VCB > 0
CUT-OFF
ACTIVE
• IC ≈ 0
• IC = IS exp(VBE /VT)
• IB ≈ 0
• IC = β IB
• VBE ≈ 0.7 V if I C non-negligible
VBE < 0
VBE > 0
REVERSE-ACTIVE
SATURATION
• IC < β IB
• VBE ≈ 0.7 V if I B non-negligible
• VCE < VBE (by definition)
VCB < 0
• Also
IE = IB + IC (always)
• THIS TABLE IS IMPORTANT - GET TO KNOW IT !
• For PNP table:
• Reverse order of suffices on all voltages in table
i.e. VCB → VBC etc
• Reverse arrows on currents in circuit
i.e. arrows on IB, IC point out of PNP device, while arrow on IE
points in.
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Common-Emitter Amplifier
Conceptual Circuit
RC
IC
V CC
VOUT
V IN
• Assume active mode:
IC = IS exp(VIN/VT)
• Apply Ohm’s Law and KVL to output side:
VOUT = VCC - RCIC
(1.3)
= VCC - RCIS exp(VIN/VT)
NOTE: Called ‘common-emitter’ because emitter is connected to
reference point for both input and output circuits. Common-Base
and Common-Collector also important.
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
C-E Amplifier
Input-Output Relationship
• e.g. VCC = 20 V, RC = 10 kΩ, IS = 10-14 A, VT = 25 mV.
VOUT (V)
20
ΔVIN
ΔVOUT
15
10
Operating Point
5
0
0.50
VIN (V)
0.55
0.60
0.65
0.70
• Plenty of voltage gain i.e. ΔVOUT >> ΔVIN
BUT:
• Highly non-linear
⇒ Output distorted unless input signal very small
⇒ Need to BIAS transistor to operate in correct region of graph
to get high gain without distortion
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
C-E Amplifier
Small-Signal Response - 1
Aim:
to get quantitative information about the small-signal voltage gain
and the linearity of a C-E amplifier
• Start with the large signal equations:
VOUT = VCC - RCIC
= VCC - RC IS exp(VIN/VT)
• Suppose we add to VIN a small input signal voltage vin, resulting in a
corresponding signal vout at the output. We can relate vout to vin by
expanding the above as a Taylor series:
VOUT + vout = VCC - RC IC [1 + vin/VT + (vin/VT)2/2 + ..]
(1.5)
• Assuming vin << VT, we can neglect quadratic and higher terms, giving:
VOUT + vout ≈ VCC - RCIC - RC(IC/VT)vin
vin << VT
This is a LINEAR APPROXIMATION, valid only when vin is small
Cont’d . .
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
C-E Amplifier
Small-Signal Response - 2
• Using (1.3), we can separate the output voltage into BIAS and SIGNAL
components:
VOUT = VCC - RCIC
vout ≈ - RC(IC /VT)vin
Quiescent O/P Voltage
Output Signal
• SMALL-SIGNAL VOLTAGE GAIN:
Av = vout/vin = - RCIC /VT = - RC gm
(1.10)
e.g. If quiescent O/P voltage lies roughly mid-way between the supply
rails then RCIC ≈ VCC /2. In this case Av = -VCC /(2VT), so for VCC = 20 V
we get AV = -400.
The quantity gm= IC/VT is known as the TRANSCONDUCTANCE of the
transistor.
• LINEARITY
Include higher order terms from Equation 1.5:
vout ≈ - Rc gm [ vin + vin 2/2 VT + . . . . ]
Ratio of unwanted quadratic term to linear term is vin/2VT,
10 % distortion when vin/2VT ≈ 0.1, or vin ≈ 5 mV.
so expect
⇒ Amplifier is linear only for very small signals
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Bias Stabilisation - 1
• Biasing at constant VBE is a bad idea, because IS and VT both vary with
temperature, and we require constant IC (or IE) for stable operation.
Also, IS is not a well-defined transistor parameter.
• We can obtain approximately constant IE as follows:
VCC
RC
VBIAS
VOUT + vout
vin
L
RE
(a)
• KVL in loop L (with no signal) gives:
IE = (VBIAS - VBE) /RE
≈ (VBIAS - 0.7 V) /RE
(1.11)
if VBIAS >> VBE
⇒ IE relatively insensitive to exact value of VBE
• Get IC from
IC = α IE
where α = β/(1 + β) ≈ 1
• α is the COMMON-BASE CURRENT GAIN
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Bias Stabilisation - 2
• RE provides NEGATIVE FEEDBACK
i.e. if the emitter current starts to rise as a result of some change in
the transistor’s characteristics, then the voltage across RE rises
accordingly. This in turn lowers the base-emitter voltage of the
transistor, tending to bring the emitter current back down towards
its original value.
⇒ STABILISATION
BUT RE also:
• Reduces small-signal voltage gain:
Av = - RC gm /(1 + IERE/VT)
(1.12)
≈ - α RC/RE
• Reduces output swing
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Bias Stabilisation - 3
Recovery of Small-Signal Voltage Gain
• We can recover the original value of Av for AC signals by using a
BYPASS CAPACITOR:
VCC
RC
V BIAS
vin
VOUT + vout
RE
CE
(b)
• Now we have:
Av = - RC gm /(1 + IEZE/VT)
(1.12b)
where ZE is the combined impedance of RE and CE:
ZE = RE /(1 + jωRECE)
By making CE large enough, we can make the parallel combination appear
like a short circuit (i.e. | ZE | ≈ 0) at all AC frequencies of interest, so that
Equation 1.12b reduces to Av ≈ - RCgm as for our original common-emitter
amplifier. On the other hand, the capacitor has no effect on biasing,
because it passes no DC current.
NB
Technique only really relevant to discrete circuits (no big capacitors
inside IC’s!)
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Example 1
Analyze the circuit below to determine the voltages at all nodes and the
currents in all branches. Assume β = 100.
1. VBE is around 0.7V
2.
3.
4.
5.
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Example 2
Analyze the circuit below to determine the voltages at all nodes and the
currents in all branches. Assume β = 100.
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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Aero2 Signals & Systems (Part 2)
Notes on BJT and transistor circuits
Step 1: Simplify base circuit using Thévenin’s theorem.
Step 2: Evaluate the base or emitter current by writing a loop equation
around the loop marked L.
• Step 3: Now evaluate all the voltages.
VB = VBE + I E RE = 0.7 + 1.29 × 3 = 4.57V
IC = (
β
1+ β
) I E = 0.99 × 1.29 = 1.28mA
VC = +15 − I C RC = 15 − 1.28 × 5 = 8.6V
(Based on Dr Holmes’ notes for EE1/ISE1 course)
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